# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Bowditch Table 15**

**From:**Nels Tomlinson

**Date:**2005 Jan 24, 11:14 -0900

Trevor, here's my interpretation of Table 15: The table is for use only when the base of the distant object is obscured by the horizon. Enter the table with the measured, corrected angle and the difference in heights, and read off the amount to add to your Dh. That's the difference between your guess and mine: you guessed it was distance to object, I'm guessing distance from your horizon to object. Notation: Dh= distance to YOUR horizon, DH =distance to object's horizon. Big H= object's height, h=your eye height. If the angle is zero, you are at the maximum, or geographic range (GR), which is given in Table 13 as GR=1.17(sqrt(H)+sqrt(h)). As the angle increases, you must be closer, and therefore the number to add must be getting smaller. The table shows that. Here's a thought experiment: for a fixed height of eye, as you stand on the beach and watch a tall ship sail toward you (your height of eye is 10 feet, the ship 100 feet tall, so H-h=90). When the angle is 0 (ship just in sight), you know that the ship must be 1.17(sqrt(10)+sqrt(100))=15.4 miles off, from Table 13. Enter table 15 with angle=0 and H-h=90 and read 11.1, add your Dh=1.17sqrt(10)=3.7 to get 14.8. Close to Table 13, but maybe not close enough. When the angle gets to 30', enter the table to read 1.7, meaning that the ship is only 1.7 miles past your horizon. It is Dh+1.7=5.4 miles off. Table 15 runs out of numbers when the angle =1 50', and the ship is 0.5 miles beyond your horizon. Here's another thought experiment: start at 30 feet up the mast so that H-h=70 and measure, say, 0 10', indicating that the ship is 3.5 miles beyond your 6.4 NM horizon. Total distance =9.9 NM. Now you climb to 70 feet, and H-h=30. Your horizon is now 9.8NM, the ship is still 9.9NM off, so the angle you measure must be larger, and in fact must be greater than 1 50', since the ship is only 0.1 mile beyond your horizon. Now, let's repeat your thought experiment. It's just like mine, except I started with an angle larger than zero. Start at 30 feet up the mast so that H-h=70 and measure 0 degrees, indicating that the ship is 6.4 miles beyond your 6.4 NM horizon, for a distance of 12.8 NM. (Unfortunately, Table 13 gives 1.17(sqrt(30)+sqrt(100))=18.1 NM) Climb to 70 feet, and the ship must now be 3 NM beyond your new horizon of 9.8 NM, suggesting that your sextant should be reading 0 15'. As you go higher up the mast, the angle gets larger, as one would expect. The problem is that Table 13 gives a ``slightly'' larger geographic distance than Table 15 does for the angle=0 case. Trevor, I think this is the objection you show in your figure 4 on that web page? Other than that, I think that my interpretation seems to make sense. Angle increases and distance beyond your horizon decreases as you climb the mast, which is satisfying. Furthermore, this interpretation deals with the fact that, for an angle of zero, you are _much_ farther off from an object 30 feet above you when you are at 1000 feet altitude than when you are at 70 feet. There are some holes left in my understanding. Here's one problem: When I try computing the formula for a few numbers, I get very different results than are shown in Table 15. Degrees H-h D from Formula D from Table 15 0 30 40.8 6.4 0 70 95.25 9.7 0.5 30 40.8 2.8 -0 4' 30 50.4 12.8 Here is an interesting point about the formula: as presented in the 2002 Bowditch, the angle only matters if it is negative. When \alpha is positive, \tan(\alpha) is positive. In the first term, squaring and then taking the square root merely takes the absolute value (i.e., throws away the sign), so if \alpha is in the range 0-90 degrees (where I think it will always be in practice), the first term cancels with the third term, and we're left with the second term: (H-h)/constant. Either I've completely misunderstood things, or there is a series of long standing blunders in Bowditch. I'm betting on the former. I'm looking forward to seeing what some of the knowledgeable list members have to say. Nels On Sun, 23 Jan 2005 00:36:43 -0400, Trevor J. Kenchingtonwrote: > I am gradually working my way through Bowditch (1995 edition), trying to > be sure that I understand everything that is relevant to smaller vessels > that the book is really intended for. Most I can follow, sometimes with > a bit of effort, but I am having trouble with Table 15. Since the table > numbers have changed between some editions, Table 15 in 1995 was the one > providing "Distance by Vertical Angle Measured Between Sea Horizon and > Top of Object Beyond Sea Horizon". (That was Table 9 in at least one > edition of Bowditch.) > > The text description accompanying the table says that it provides the > distance to the object (by implication: from the observer to the > object), given a knowledge of the difference in height between the > observer's eye and the object, plus a vertical angle measured between > the top of the object and the horizon (that angle being corrected for IE > and dip). [The Bowditch main text doesn't seem to refer to Table 15 and > mixes this business of an object beyond the horizon with the common > business of distance off by vertical angle of an object of known height > whose waterline is visible.] > > Table 15 itself does give distances from observer #1 to the object > observed, if the observer #1 has zero height of eye and the top of the > object is just dipping below the horizon. At least, the tabulated > distance then equals the horizon distance for an observer (#2) at the > top of the object, which should be the same thing. > > However, if Table 15 was to be believed, as observer #1 climbs the mast > of his boat and increases his height of eye, the object observed would > drop below his horizon (assuming its true distance did not change), > which is obviously nonsensical. > > So ... if Table 15 does not give the distance from observer to object, > except in special cases, what does it provide? > > I'm guessing that it might be the distance from the object to the > observer's horizon, to which the observer's horizon distance must be > added to get the full distance from observer to object. However, I do > not know of any way to either confirm of refute that idea (save for a > full-scale experiment, which is a bit beyond my resources just now). > > Can anyone sort this out for me? > > Trevor Kenchington > > -- > Trevor J. Kenchington PhD Gadus{at}iStar.ca > Gadus Associates, Office(902) 889-9250 > R.R.#1, Musquodoboit Harbour, Fax (902) 889-9251 > Nova Scotia B0J 2L0, CANADA Home (902) 889-3555 > > Science Serving the Fisheries > http://home.istar.ca/~gadus >